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Unformatted text preview: CHAPTER 12 Partial Differential Equations (PDEs) SECTION 12.1 Basic Concepts, page 535 Purpose. To familiarize the student with the following: Concept of solution, verification of solutions Superposition principle for homogeneous linear PDEs PDEs solvable by methods for ODEs SOLUTIONS TO PROBLEM SET 12.1, page 537 2. u 5 c 1 ( y ) e x 1 c 2 ( y ) e 2 x . Problems 1–12 will help the student get used to the notations in this chapter; in particular, y will now occur as an independent variable. Second-order PDEs in this set will also help review the solution methods in Chap. 2, which will play a role in separating variables. 4. u 5 c ( x ) e 2 y 2 6. u 5 c 1 ( y ) e 2 2 xy 1 c 2 ( y ) e 2 xy 8. ln u 5 2 x E y dy 5 xy 2 1 c | ( x ), u 5 c ( x ) e xy 2 10. Set u y 5 v . Then v y 5 4 x v , v 5 c | 1 ( x ) e 4 xy ; hence u 5 E v dy 5 c 1 ( x ) e 4 xy 1 c 2 ( x ). 12. u 5 ( c 1 ( x ) 1 c 2 ( x ) y ) e 2 5 y 1 1 _ 2 y 2 e 2 5 y . The function on the right is a solution of the homogeneous ODE, corresponding to a double root, so that the last term in the solution involves the factor y 2 . 14. c 5 1/2. Problems 14–25 should give the student a first impression of what kind of solutions to expect, and of the great variety of solutions compared with those of ODEs. It should be emphasized that although the wave and the heat equations look so similar, their solutions are basically different. It could be mentioned that the boundary and initial conditions are basically different, too. Of course, this will be seen in great detail in later sections, so one should perhaps be cautious not to overload students with such details before they have seen a problem being solved. 16. c 5 6 18. c 5 Ï k /32 w 20. c 5 2, v arbitrary 22. Solutions of the Laplace equation in two dimensions will be derived systematically in complex analysis. Nevertheless, it may be useful to see an unsystematic selection of typical solutions, as given in (7) and in Probs. 23–25. 26. Team Project. (a) Denoting derivatives with respect to the entire argument x 1 ct and x 2 ct , respectively, by a prime, we obtain by differentiating twice u xx 5 v 0 1 w , u tt 5 v c 2 1 w c 2 and from this the desired result. 228 im12.qxd 9/21/05 5:16 PM Page 228 (c) The student should realize that u 5 1/ Ï x 2 1 y w 2 w is not a solution of Laplace’s equation in two variables. It satisfies the Poisson equation with ƒ 5 ( x 2 1 y 2 ) 2 3/2 , which seems remarkable. 28. A function whose first partial derivatives are zero is a constant, u ( x , y ) 5 c 5 const. Integrate the first PDE and then use the second. 30. Integrating the first PDE and the second PDE gives u 5 c 1 ( y ) x 1 c 2 ( y ) and u 5 c 3 ( x ) y 1 c 4 ( x ), respectively. Equating these two functions gives u 5 axy 1 bx 1 cy 1 k ....
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